Answer
(a) $d = 0.51~m$
(b) $d = 0.38~m$
Work Step by Step
(a) We can use conservation of energy to find the maximum height:
$U_g = U_e$
$mgh = \frac{1}{2}kx^2$
$h = \frac{kx^2}{2mg}$
$h = \frac{(25~N/m)(0.10~m)^2}{(2)(0.050~kg)(9.8~m/s^2)}$
$h = 0.255~m$
We can find the distance up the slope:
$\frac{h}{d} = sin~\theta$
$d = \frac{h}{sin~\theta}$
$d = \frac{0.255~m}{sin~30^{\circ}}$
$d = 0.51~m$
(b) We can use work and energy to find the distance traveled along the slope:
$U_g = U_e+W_{ext}$
$mgh = \frac{1}{2}kx^2-mg~cos~\theta~\mu_k~d$
$mg~d~sin~\theta + mg~cos~\theta~\mu_k~d= \frac{1}{2}kx^2$
$d = \frac{kx^2}{(2)(mg~sin~\theta + mg~cos~\theta~\mu_k)}$
$d = \frac{(25~N/m)(0.10~m)^2}{(2)(0.050~kg)(9.8~m/s^2)(sin~30^{\circ}+0.20~cos~30^{\circ})}$
$d = 0.38~m$
